Example of use of Adiabatic Equation.

[tecmatt]

Hi guys
needs some help. been a while since iv done this and seem to be having a blonde moment.
trying to work out the min csa of the cpc on a 3 phase 20A circuit. currently the cpc is only 2.5mm. Ze=0.05 R1+R2=0.23 k=115.
Id appreciate any advise/help, I keep getting odd results and other people iv spoke to are in the same position of not doing it for years

 

[Guest123]

Hi guys
needs some help. been a while since iv done this and seem to be having a blonde moment.
trying to work out the min csa of the cpc on a 3 phase 20A circuit. currently the cpc is only 2.5mm. Ze=0.05 R1+R2=0.23 k=115.
Id appreciate any advise/help, I keep getting odd results and other people iv spoke to are in the same position of not doing it for years

Ok so circuit Zs = 0.05 + 0.23 = 0.28ohms.

Fault current = 230 / 0.28 = 821.4A (I)

I assume the protective to be a 60898 'B', with a fault current of that value the device will operate in 0.1 so (T) = 0.1

Now we have (I) (T) & (K)

S = Sqroot of (I squared x T) / K

I squared = 674697.9

674697.9 x 0.1 = 67469.7

Sqroot of 67469.7 = 259.7

259.7 / 115 = 2.25, nearest available size = 2.5mm.

 

[redface]

so im trying to work out the minimum size main earth for main supply ze=0.11 its a tncs supply i would say k=143 and for the main consumerunit i would say its 5 second disconection time so
230/0.11=2091
so doing the caculation i get 32.9mm and this cant be right what am i doing wrong it must be T
a bs 1361 will go quicker than 0.1 s at that fault curent.
so taking pg 244 time/current characteristics for 1361 60 a fuse at 5 seconds is 330A so doing the calc again i get 5.16mm this is better but what is right or for a consumerunit disconection time is it 0.4
please help i done it on a pir with a 6mm main earth which i doubled up to 12mm but minimum size is 5.16mm now this is safe just not conforming to bs7671 2008

 

[andyb]

so im trying to work out the minimum size main earth for main supply ze=0.11 its a tncs supply i would say k=143 and for the main consumerunit i would say its 5 second disconection time so
230/0.11=2091
so doing the caculation i get 32.9mm and this cant be right what am i doing wrong it must be T
a bs 1361 will go quicker than 0.1 s at that fault curent.
so taking pg 244 time/current characteristics for 1361 60 a fuse at 5 seconds is 330A so doing the calc again i get 5.16mm this is better but what is right or for a consumerunit disconection time is it 0.4
please help i done it on a pir with a 6mm main earth which i doubled up to 12mm but minimum size is 5.16mm now this is safe just not conforming to bs7671 2008

I think it must be your maths

As you say, fault current of 2091
so S=sq rt ((2091x2091)0.1)/143
=sq rt(4372281 x 0.1)/143
=sq rt437228/143
=661.23/143

S=4.6mm

Don't forget though that the main bonding conductor will need to be 10mm.

 

[Mr CJ]

Many thanks Dave.

 

[sinemon]

However he did add it in his working out and not multiply (0.08+0.037)

 

[brizospark]

where do i find k values?

 

[topquark]

Table 43.1 in BRB/BGB (page 77 in BRB, page 85 in BGB).

 

[brizospark]

what do k values stand 4?

 

[topquark]

Resistivity of the material the conductor is made of, taking into account temperature and heat capacity of the conductor material.

 

[topquark]

Resistivity of the material the conductor is made of, taking into account temperature and heat capacity of the conductor material.

There is a more complete (and technically accurate) definition just above table 43.1.

 

[ian65]

Very good example thanks for sharig it i am due to take my exam next month, and i am told it is 50/50 if this question comes up

\thanks once again Ian.

 

[MacSparky]

Its been a while but I have my NAPIT inspection tuesday week and jumped on here for a refresher....lovely example of the adiabatic and easy to understand when put the way it was...thanks Dave....i'll keep reading....best sparky forum IMO.

 

[SIIVION]

I too have just joined AEI Cables. Thanks for the tip AEI has some very useful info.

 

[pactrol]

Hi If I may chip in what you say is correct but the driving force behind pme is to save the REC money on not having to provide an earth conductor .
my view on pme is its dangerous as it relys on the consumer having to install larger than normal earthing conductors in case the pme neutral fails otherwise you have a very dangerous situation give me TN-S any day

 

[tony mc]

If the PME neutral was to fail then no matter what size earthing conductor you had fitted the earth would still fail as the return path for the earth is the neutral. (Main bonding conductors are sized according to the neutral)

I was under the impression that if the DNO did not supply a earth connection then it was down to the customer who would then have to go down the TT route.

There have been a few posts recently re high ZE readings on TNS systems which turned out to be the suppliers sheath broken!

 

[jojerome12]

yes thanks for that,

 

[jojerome12]

that is a great help, always need to be aware, especially when you have not been doing it for a awhile

 

[danny26]

wqesasa

 

[pactrol]

Hi Tony mc

Im sorry I have to disagree with you pme installations have multiple earth spikes as back up for just such a senario they usually fit a earth rod at the transformer & another where the service enters the building so loss of a neutral will divert to the earthing system back to the transformer hence the need for large earth wires & bonding as every customer fed of that transformer has a neutral return via the earth. ( very dodgy but the rec don,t follow Bs7671 )